mechanics of fluids If force, length and the time are selected as the three fund
ID: 1717447 • Letter: M
Question
mechanics of fluids If force, length and the time are selected as the three fundamental dimensions, the units of mass in the SI system could be written as Develop the dimension of kinetic viscosity (v) using the F-L-T system A 3600N force acts on a 250cm^2 area at an angle of 60degree to the normal. The pressure (normal stress) acting on the area is (show your result in Pa): A liquid with a specific gravity of 1.2 fills a volume. If the mass in the volume 10 slug, What is the magnitude of the volume in ft^1? The mass of a car is 3.2 tom. which are evenly distributed to 4 tires, for a measured tire pressure of 32 psi Suppose the local air pressure is 100 kpa, compute the absolute pressure inside foe tire. Compute the contacting area between each tire and the ground using SI system.Explanation / Answer
>> Fundamental Dimensions are :- Force (F), Length (L) and Time (T)
>> Now, for mass (m), As, F = ma
As, acceleration, a = v/t = L/t2
=> m = F/a
=> m = F*t2/L
=> Dimensions of mass are:- [F T2 L-1] ANSWER..................(1).......
>> Solution (2)
>> Kinematic Viscosity, v
>> As, Stress, F/A = (v*d)*(U/L) ( Newton's Equation).......(2)
, As, here, F = Force,
A = Area = [L2]
v = Kinematic Viscosity,
d = density = mass/volume = [F T2 L-1]/L3 ( From (1) )....
=> d = [ F T2 L-4 ]
>> U = Velocity = L/T = [ L T-1 ]
>> Now, putting all in basic equation,(2).
=> [F] [L-2] = v [ F T2 L-4 ] [ L T-1 ] [ L-1 ]
=> v = [ L2 T-1 ] ........ANSWER.....DIMENSIONS OF KINEMATIC VISCOSITY.......
>> Solution (3).
>> Force, F = 3600 N ,
Force, F acts at an angle with normal of cross section, and = 60 degree
>> A = 250 cm2 = 0.025 m2
>> So,component of force along normal to the surface Fn = F*cos60 = 1800 N
=> Normal Stress = Fn/A = 1800/0.025 = 720 KPa ..........ANSWER.........
>> Solution (4).
>>
mass = 10 slug = 321.74 lb ( bcz 1 slg = 32.174 lb)
>> Specific Gravity = 1.2
=> d = density of fluid = 1.2*1000 = 1200 Kg/m3 = 1200*0.06243 lb/ft3 = 74.916 lb/ft3 [ 1 kg/m3 = 0.06243 lb/ft3 ]
>> As, density = mass/volume
=> Volume, V = m/d = 321.74/74.916 = 4.295 ft3 ............ANSWER........
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